45.Semiconductor and semiconductor devices by Aditya Bhardwaj

CHAPTER - 45

SEMICONDUCTOR AND SEMICONDUCTOR DEVICES 1.

f = 1013 kg/m , V = 1 m m = fV = 1013  1 = 1013 kg

1013  103  6  1023 26 = 264.26  10 . 23 26 28 26 a) Total no.of states = 2 N = 2  264.26  10 = 528.52 = 5.3  10  10 26 b) Total no.of unoccupied states = 2.65  10 . In a pure semiconductor, the no.of conduction electrons = no.of holes Given volume = 1 cm  1 cm  1 mm –2 –2 –3 –7 3 = 1  10  1  10  1  10 = 10 m 19 –7 12 No.of electrons = 6  10  10 = 6  10 . 12 Hence no.of holes = 6  10 . –23 E = 0.23 eV, K = 1.38  10 KT = E –23 –19  1.38  10  T = 0.23  1.6  10 No.of atoms =

0.23  1.6  10 19

0.23  1.6  10 4 4 = 0.2676  10 = 2670. 1.38 1.38  10 Bandgap = 1.1 eV, T = 300 K T=

a) Ratio =

23

1.1 1.1  = 42.53= 43 KT 8.62  10 5  3  102

1.1 105 = 3000.47 K. 4.253  8.62 8.62  105  T 2KT = Energy gap between acceptor band and valency band –23  2  1.38  10  300 b) 4.253 =



1.1

or T =

6  1.38 10 21  6  1.38  2  19 eV     10 eV 1.6 10  1.6  –2 = 5.175  10 eV = 51.75 meV = 50 meV. Given : Band gap = 3.2 eV, E = hc /  = 1242 /  = 3.2 or  = 388.1 nm.  = 820 nm E = hc /  = 1242/820 = 1.5 eV Band Gap = 0.65 eV,  =? –9 –5 E = hc /  = 1242 / 0.65 = 1910.7  10 m = 1.9  10 m. Band gap = Energy need to over come the gap  E = (2  1.38  3)  10

7. 8. 9.

–21

hc 1242eV  nm = 2.0 eV.   620nm 10. Given n = e E / 2KT , E = Diamon  6 eV ; E Si  1.1 eV 6

Now, n1 = e

E1 / 2KT



5 e 23008.6210

1.1

n2 = e E2 / 2KT  e 23008.6210

5

n1 4.14772  1051 –42  = 7.15  10 . 10 n2 5.7978  10 Due to more E, the conduction electrons per cubic metre in diamond is almost zero. 1

Semiconductor devices 11.  = T

3/2

eE / 2KT at 4°K 0.74

 = 43 / 2  e 28.6210 At 300 K,

5

4

=8e

0.67

 = 300 Ratio =

3/2

5 e 28.6210 300



–1073.08

3  1730 12.95 . e 8

8  e 1073.08 [(3  1730) / 8]  e 12.95

64 e 1060.13 . 3  1730 15

12. Total no.of charge carriers initially = 2  7  10 = 14  10 /Cubic meter 17 3 Finally the total no.of charge carriers = 14  10 / m We know : The product of the concentrations of holes and conduction electrons remains, almost the same. Let x be the no.of holes. 15 15 17 So, (7  10 )  (7  10 ) = x  (14  10 – x) 17 2 30  14x  10 – x = 79  10 2 17 30  x – 14x  10 – 49  10 = 0

14  1017  142  1034  4  49  1030 17 = 14.00035  10 . 2 = Increased in no.of holes or the no.of atoms of Boron added. x=

5  1028

–3 13 10 = 3.607  10  10 = 3.607  10 . 1386.035  1015 13. (No. of holes) (No.of conduction electrons) = constant. At first : 19 No. of conduction electrons = 6  10 19 No.of holes = 6  10 After doping 23 No.of conduction electrons = 2  10 No. of holes = x. 19 19 23 (6  10 ) (6  10 ) = (2  10 )x

 1 atom of Boron is added per



6  6  1019 19

=x 2  1023 15 16  x = 18  10 = 1.8  10 . 14.  = 0 eE / 2KT E = 0.650 eV, T = 300 K –5 According to question, K = 8.62  10 eV E

0 eE / 2KT  2  0 e 2K300 0.65 5

–5

 e 28.6210 T = 6.96561  10 Taking in on both sides, We get, 

0.65 2  8.62  10 5  T '

= –11.874525

1 11.574525  2  8.62  10 5  T' 0.65

 T’ = 317.51178 = 318 K.

Semiconductor devices 15. Given band gap = 1 eV –3 Net band gap after doping = (1 – 10 )eV = 0.999 eV According to the question, KT1 = 0.999/50  T1 = 231.78 = 231.8 For the maximum limit KT2 = 2  0.999

2  1 10 3

2  102  23.2 . 8.62 8.62  10 Temperature range is (23.2 – 231.8). –7 Depletion region ‘d’ = 400 nm = 4  10 m 5 Electric field E = 5  10 V/m a) Potential barrier V = E  d = 0.2 V b) Kinetic energy required = Potential barrier  e = 0.2 eV [Where e = Charge of electron] Potential barrier = 0.2 Volt n a) K.E. = (Potential difference)  e = 0.2 eV (in unbiased cond ) b) In forward biasing KE + Ve = 0.2e  KE = 0.2e – 0.1e = 0.1e. c) In reverse biasing KE – Ve = 0.2 e  KE = 0.2e + 0.1e = 0.3e. Potential barrier ‘d’ = 250 meV Initial KE of hole = 300 meV We know : KE of the hole decreases when the junction is forward biased and increases when reverse blased in the given ‘Pn’ diode. So, a) Final KE = (300 – 250) meV = 50 meV b) Initial KE = (300 + 250) meV = 550 meV i1 = 25 A, V = 200 mV, i2 = 75 A a) When in unbiased condition drift current = diffusion current  Diffusion current = 25 A. b) On reverse biasing the diffusion current becomes ‘O’. c) On forward biasing the actual current be x. x – Drift current = Forward biasing current  x – 25 A = 75 A  x = (75 + 25) A = 100 A. –6 Drift current = 20 A = 20  10 A. Both holes and electrons are moving  T2 =

16.

17.

18.

19.

20.

5



So, no.of electrons = aV/KT

21. a) e

20  106 2  1.6  10 19

= 6.25  10 .

= 100 V

  R=

5 e 8.6210 300

V 8.6210 5 300

= 100

= 4.605  V = 4.605  8.62  3  10

–3

= 119.08  10

–3

V V 119.08  10 3 119.08  10 3 2   = = 1.2  10 . I I0 (eev / KT 1 ) 10  10 6  (100  1) 99  10 5

V0 = I0R -6 2 –3  10  10  1.2  10 = 1.2  10 = 0.0012 V. 3

Semiconductor devices c) 0.2 =

KT  eV / KT e ei0 –5

K = 8.62  10 eV/K, T = 300 K –5 i0 = 10  10 A. Substituting the values in the equation and solving We get V = 0.25 –6 22. a) i0 = 20  10 A, T = 300 K, V = 300 mV ev

i = i0 e KT

1

100

= 20  10 6 (e 8.62  1) = 2.18 A = 2 A. V

b) 4 = 20  10 

V 103 8.62 e 3

6

2 (e 8.62310

 200001 

 1) 

V 103 8.62 e 3

1

4  106 20

V 103  12.2060 8.623

 V = 315 mV = 318 mV. 23. a) Current in the circuit = Drift current (Since, the diode is reverse biased = 20 A) –6 b) Voltage across the diode = 5 – (20  20  10 ) -4 = 5 – (4  10 ) = 5 V. 24. From the figure : According to wheat stone bridge principle, there is no current through the diode.

5

A

40 = 20 . Hence net resistance of the circuit is 2 25. a) Since both the diodes are forward biased net resistance = 0 2V =1A i= 2

20

B

i

2

b) One of the diodes is forward biased and other is reverse biase. Thus the resistance of one becomes . i=

2 = 0 A. 2

Both are forward biased. Thus the resistance is 0. i=

2 = 1 A. 2

One is forward biased and other is reverse biased. Thus the current passes through the forward biased diode. 2 i= = 1 A. 2 26. The diode is reverse biased. Hence the resistance is infinite. So, current through A1 is zero. For A2, current =

2 = 0.2 Amp. 10

A1 2V A2

10

Semiconductor devices i 10

27. Both diodes are forward biased. Thus the net diode resistance is 0. i=

5 5  = 1 A. (10  10) /10.10 5

i 5V

One diode is forward biased and other is reverse biased.

i

Current passes through the forward biased diode only. i=

V 5 = 1/2 = 0.5 A.  Rnet 10  0

28. a) When R = 12  The wire EF becomes ineffective due to the net (–)ve voltage. Hence, current through R = 10/24 = 0.4166 = 0.42 A. b) Similarly for R = 48 . 

A

C E

10 = 10/60 = 0.16 A. (48  12)

10 10

10

12

R 1

B D F

29. A

10

B

X

V

i

V

1 2

A

B

Since the diode 2 is reverse biased no current will pass through it. I

X

V

30. Let the potentials at A and B be VA and VB respectively. i) If VA > VB Then current flows from A to B and the diode is in forward biased. Eq. Resistance = 10/2 = 5 . ii) If VA < VB  Then current flows from B to A and the diode is reverse biased. Hence Eq.Resistance = 10 . –6 31. Ib = 80 A – 30 A = 50 A = 50  10 A –3 Ic = 3.5 mA – 1 mA = –2.5 mA = 2.5  10 A

 I =  c  Ib

  Vce = constant 

2.5  10 3

2500 = 50. 50 50  10 Current gain = 50. 

6



A

B

Semiconductor devices 32.  = 50, Ib = 50 A, V0 =   RG = 50  2/0.5 = 200. V V0 200  = 8000 V. a) VG = V0/V1 = 0  Vi Ib  Ri 50  106  5  102 -6

b) Vi = Ib  Ri = 50  10  5  10 = 0.00025 V = 25 mV. R 2 2 4 = 10 . c) Power gain =   RG = 2  0  2500  Ri 0.5 33. X = ABC  BCA  CAB a) A = 1, B = 0, C = 1 X = 1. b) A = B = C = 1 X = 0. 34. For ABC  BCA B C A

35. LHS = AB  AB = X  X

[X = AB]

If X = 0, X = 1 If X = 0, X = 1  1 + 0 or 0 + 1 = 1  RHS = 1 (Proved)

  `

C

P R1=0.5K 

2K

N N e